Question

我正在开发纸牌游戏。我的游戏中有一张纸牌目录，比如说“目录”表：

class Sample extends StatelessWidget {
  @override
  Widget build(BuildContext context) {
    return Stack(
      children: <Widget>[
        Column(
          children: <Widget>[
            Container(
              height: MediaQuery.of(context).size.height / 2,
              width: double.infinity,
              decoration: BoxDecoration(
                color: Color.fromRGBO(50, 50, 205, 0.3),
                image: DecorationImage(
                  image: NetworkImage(
                    'https://cdn.pixabay.com/photo/2015/04/23/21/36/auto-736794__340.jpg',
                  ),
                  fit: BoxFit.cover,
                ),
              ),
            ),
          ],
        ),
        Scaffold(
          backgroundColor: Colors.transparent,
          drawer: Drawer(),
          appBar: AppBar(
            backgroundColor: Colors.transparent,
            elevation: 0.0,
            title: new Text("Demo"),
            centerTitle: true,
          ),
          body: Column(
            children: <Widget>[
              Expanded(
                child: Center(
                  child: new Container(
                    decoration: BoxDecoration(
                      color: Colors.red
                    ),
                    child: new Text("book"),
                  ),
                ),
              )
            ],
          ),
        ),
      ],
    );
  }
}

在另一个桌子上（名称为playerCards），我拥有一个玩家拥有的所有卡牌。在这种结构上，我不想重复我为目录卡设置的所有信息。我只需要存储卡的ID和金额：

[
  {
    "id": 1,
    "name": "my spell",
    "type": "spell",
    "cost": 120
  }
]

我想知道如何查询以以下格式退回玩家的所有卡牌：

[
  {
    "id": 1,
    "card_count_in_book": 10,
    "card_count_in_deck": 1
  }
]

Answer 1

使用$lookup运算符通过id合并两个集合中的数据

db.Catalog.aggregate([
    {
        $lookup: {
            from: "playerCards",
            localField: "id",
            foreignField: "id",
            as: "playercards"
        }
    },
    {
        $unwind: "$playercards"
    },
    {
        $project: {
            id: 1,
            name: 1,
            type: 1,
            cost: 1,
            card_count_in_book: "$playercards.card_count_in_book",
            card_count_in_deck: "$playercards.card_count_in_deck",
        }
    }
])

mongoDB如何加入/合并结果

1 个答案: